In the neighbourhood of `x=0` it is known that `1+|x|lt(e^(x)-1)/(x)lt1-|x|"then find"lim_(xto0)(e^(x)-1)/(x).`
In the neighbourhood of `x=0` it is known that
`1+|x|lt(e^(x)-1)/(x)lt1-|x|"then find"lim_(xto0)(e^(x)-1)/(x).`
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We have `1+|x|lt(e^(x)1-)/(x)lt1-|x|"for "x in(0-delta,0+delta).`
`:." "`Using Sandwich rule,
`underset(xto0)lim(1+|x|)ltunderset(xto0)lim(e^(x)-1)/(x)ltunderset(xto0)lim(1-|x|)`
`implies" "1ltunderset(xto0)lim(e^(x)-1)/(x)lt1`
`underset(xto0)lim(e^(x)-1)/(x)=1`
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